Rotational Motion — Page 6 | JEE Physics

Q51

If

\vec{L}

and

\vec{P}

represent the angular momentum and linear momentum respectively of a particle of mass '

m

' having position vector as

\vec{r}=a(\hat{i} \cos \omega t+\hat{j} \sin \omega t)

. The direction of force is

A Opposite to the direction of

\vec{L}

B Opposite to the direction of

\vec{L} \times \vec{P}

C Opposite to the direction of

\vec{r}

D Opposite to the direction of

\vec{P}

Correct Answer

Option C

Solution

Let's analyze the situation step by step. The particle’s position vector is given by

\vec{r} = a(\hat{i}\cos\omega t + \hat{j}\sin\omega t),

which represents uniform circular motion in the xy-plane. Its velocity is the time derivative of the position:

\vec{v} = \frac{d\vec{r}}{dt} = -a\omega\sin\omega t\,\hat{i} + a\omega\cos\omega t\,\hat{j}.

The acceleration, obtained by differentiating the velocity, is:

\vec{a} = \frac{d\vec{v}}{dt} = -a\omega^2\cos\omega t\,\hat{i} - a\omega^2\sin\omega t\,\hat{j}.

Notice that this can be rewritten as:

\vec{a} = -\omega^2 \vec{r}.

This indicates that the acceleration (and hence the force, since

\vec{F}=m\vec{a}

) is directed opposite to the position vector

\vec{r}

—that is, radially inward. To check against the given options: Option A (Opposite to

\vec{L}

): The angular momentum

\vec{L}=m\vec{r}\times\vec{v}

is perpendicular to the plane of motion (along

\pm\hat{k}

) and does not indicate a direction in the plane. Option B (Opposite to

\vec{L}\times\vec{P}

): For a particle in circular motion, when you compute

\vec{L}\times\vec{P}

(with

\vec{P} = m\vec{v}

), you’ll find that it is proportional to

-\vec{r}.

Thus, the direction opposite to

\vec{L}\times\vec{P}

would be the same as

\vec{r}

, which is the outward radial direction—not matching the inward (centripetal) force. Option D (Opposite to

\vec{P}

): The linear momentum is tangential; the force (centripetal) is not tangential but directed radially inward.

Option C (Opposite to

\vec{r}

): As we found, the acceleration (and hence force) is given by

\vec{F}= m\vec{a} = -m\omega^2 \vec{r},

which is clearly opposite to

\vec{r}.

Therefore, the force is directed opposite to

\vec{r},

making Option C the correct answer.

Q52

Match with M{R^2}$$

List - I		List - II
(B)	Moment of inertia of hollow sphere of radius (R) about any tangent.	(II)	${7 \over 5}M{R^2}$
(C)	Moment of inertia of circular ring of radius (R) about its diameter.	(III)	${1 \over 4}M{R^2}$
(D)	Moment of inertia of circular disc of radius (R) about any diameter.	(IV)	${1 \over 2}M{R^2}$

A A - II, B - I, C - IV, D - III

B A - I, B - II, C - IV, D - III

C A - II, B - I, C - III, D - IV

D A - I, B - II, C - III, D - IV

Correct Answer

Option A

Solution

(A) Moment of inertia of solid sphere of radius R about a tangent

= {2 \over 5}M{R^2} + M{R^2} = {7 \over 5}M{R^2}

$\Rightarrow$ A $-$ (II) (B) Moment of inertia of hollow sphere of radius R about a tangent

= {2 \over 3}M{R^2} + M{R^2} = {5 \over 3}M{R^2}

$\Rightarrow$ B $-$ (I) (C) Moment of inertia of circular ring of radius (R) about its diameter =

{{\left( {M{R^2}} \right)} \over 2}

$\Rightarrow$ C $-$ (IV) (D) Moment of inertia of circular disc of radius (R) about any diameter

= {{M{R^2}/2} \over 2} = {{M{R^2}} \over 4}

$\Rightarrow$ D $-$ (III)

Q53

One solid sphere

A

and another hollow sphere

B

are of same mass and same outer radii. Their moment of inertia about their diameters are respectively

{I_A}

and

{I_B}

such that

A

{I_A} < {I_B}

B

{I_A} > {I_B}

C

{I_A} = {I_B}

D

{{{I_A}} \over {{I_B}}} = {{{d_A}} \over {{d_B}}}

where

{d_A}

and

{d_B}

are their densities.

Correct Answer

Option A

Solution

For solid sphere the moment of inertia of

A

about its diameter

{I_A} = {2 \over 5}M{R^2}.

The moment of inertia of a hollow sphere

B

about its diameter

{I_B} = {2 \over 3}M{R^2}.

$\therefore$

{I_A} < {I_B}

Q54

A homogeneous solid cylindrical roller of radius R and mass M is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is -

A

{F \over {2mR}}

B

{2F \over {3mR}}

C

{3F \over {2mR}}

D

{F \over {3mR}}

Correct Answer

Option B

Solution

FR =

{3 \over 2}

MR2 $\alpha$ $\alpha$ =

{{2F} \over {3MR}}

Q55

The radius of gyration of a uniform rod of length

l

, about an axis passing through a point

{l \over 4}

away from the centre of the rod, and perpendicular to it, is :

A

{1 \over 8}l

B

{1 \over 4}l

C

\sqrt {{7 \over {48}}} l

D

\sqrt {{3 \over 8}} l

Correct Answer

Option C

Solution

I =

{{M{l^2}} \over {12}} + M{\left( {{l \over 4}} \right)^2}

$\Rightarrow$ I =

{{7M{l^2}} \over {48}}

$\Rightarrow$ MK2 =

{{7M{l^2}} \over {48}}

$\Rightarrow$ K =

\sqrt {{7 \over {48}}} l

Q56

A body rolls down an inclined plane without slipping. The kinetic energy of rotation is 50% of its translational kinetic energy. The body is :

A Solid sphere

B Solid cylinder

C Hollow cylinder

D Ring

Correct Answer

Option B

Solution

{1 \over 2}I{\omega ^2} = {1 \over 2} \times {1 \over 2}m{v^2}

I = {1 \over 2}m{R^2}

Body is solid cylinder

Q57

Angular momentum of a single particle moving with constant speed along circular path :

A changes in magnitude but remains same in the direction

B remains same in magnitude and direction

C remains same in magnitude but changes in the direction

D is zero

Correct Answer

Option B

Solution

\left| {\overrightarrow L } \right|

= mvr And direction will be upward & remain constant. Option (b)

Q58

A heavy iron bar of weight

12 \mathrm{~kg}

is having its one end on the ground and the other on the shoulder of a man. The rod makes an angle

60^{\circ}

with the horizontal, the weight experienced by the man is :

A

3 \mathrm{~kg}

B

6 \mathrm{~kg}

C

6 \sqrt{3} \mathrm{~kg}

D

12 \mathrm{~kg}

Correct Answer

Option A

Solution

First, let us restate the problem in our own words: We have a uniform iron bar of weight $12\,\mathrm{kgf}$ (meaning its mass is $12\,\mathrm{kg}$ , so its weight is $12g\,\mathrm{N}$ in SI units).

One end of the bar is on the ground (assume a frictionless contact), and the other end is on a man’s shoulder (also assume frictionless contact).

The bar makes an angle of $60^\circ$ with the horizontal.

We want to find how much load (“effective weight”) the man feels on his shoulder.

Under these (common) assumptions of frictionless contacts at both ends, each contact force is perpendicular to the bar.

We can solve the problem by taking torques about the ground contact point.

1.

Labeling and geometry Let the length of the bar be $L$ .

The bar is inclined at $60^\circ$ from the horizontal.

Because the bar is uniform, its center of gravity (C.O.G.) lies at its midpoint, i.e., at a distance $L/2$ from either end.

Let the weight of the bar be $W = 12g$ .

Let the reaction force from the man’s shoulder be $R$ .

(Magnitude unknown.)

Let the reaction force at the ground be $R_g$ .

(We do not actually need its exact value to answer the question.)

We will place our origin (for torque calculation) at the ground contact point.

2.

Torques about the ground contact Torque due to the bar’s weight The weight $W = 12g$ acts downward at the midpoint of the bar.

The position vector from the pivot (ground) to the bar’s midpoint is $\tfrac{L}{2}$ long and makes $60^\circ$ with the horizontal.

The weight is vertical downward, i.e., $90^\circ$ from the horizontal.

The angle between the position vector ( $60^\circ$ from horizontal) and the weight ( $90^\circ$ from horizontal) is $90^\circ - 60^\circ \;=\; 30^\circ.$ Hence, the torque magnitude from the weight is $\tau_W \;=\; \bigl(12g\bigr)\,\bigl(\tfrac{L}{2}\bigr)\,\sin(30^\circ) \;=\; 12g \cdot \tfrac{L}{2}\cdot \tfrac12 \;=\; 12g \cdot \tfrac{L}{4}.$ Torque due to the man’s reaction force $R$ The man’s shoulder is at the other end of the bar, i.e., $L$ from the pivot.

For frictionless contact, $R$ is perpendicular to the bar.

Since the bar is at $60^\circ$ from horizontal, a line perpendicular to the bar is $90^\circ$ away from the bar’s direction, so the angle between the bar’s position vector ( $\vec{r}$ ) and $R$ is $90^\circ$ .

Thus, the torque from $R$ about the ground contact is $\tau_R \;=\; R \,\times\, L \,\times\, \sin(90^\circ) \;=\; R \cdot L.$ Because the bar is in static equilibrium, the net torque about the ground contact must be zero.

Taking the clockwise direction (by weight) as positive, we have $\tau_W \;-\; \tau_R \;=\; 0 \quad\Longrightarrow\quad 12g \cdot \dfrac{L}{4} \;=\; R \cdot L.$ Solving for $R$ , $R \;=\; \dfrac{12g}{4} \;=\; 3g.$ Hence, the force on the man’s shoulder has magnitude $3g$ in newtons.

3.

Converting to “kg weight” Often, such problems phrase the answer in “kg of weight” rather than newtons.

Since $W = mg$ for a mass $m$ in Earth’s gravity, a force of $3g\,\mathrm{N}$ corresponds to a “weight” of $3\,\mathrm{kgf}$ .

So the man effectively feels 3 kg of load on his shoulder.

Q59

The moment of inertia of a circular ring of mass M and diameter r about a tangential axis lying in the plane of the ring is :

A

\dfrac{3}{8} \mathrm{Mr}^2

B

2 \mathrm{Mr}^2

C

\dfrac{1}{2} \mathrm{Mr}^2

D

\dfrac{3}{2} \mathrm{Mr}^2

Correct Answer

Option A

Solution

The moment of inertia of a circular ring with mass $M$ and diameter $r$ about a tangential axis lying in the plane of the ring is calculated as follows: Given the diameter $r$ of the ring, the radius $R$ is $\dfrac{r}{2}$ .

The formula for the moment of inertia about a tangential axis in the plane of the ring is: $I_{\text{tangential}} = \dfrac{3}{2} M \left(\dfrac{R}{2}\right)^2$ Substitute the value of $R = \dfrac{r}{2}$ into the equation: $I_{\text{tangential}} = \dfrac{3}{2} M \left(\dfrac{r}{2}\right)^2 = \dfrac{3}{8} Mr^2$ Therefore, the moment of inertia of the ring about the specified axis is $\dfrac{3}{8} Mr^2$ .

Q60

A round uniform body of radius

R,

mass

M

and moment of inertia

I

rolls down (without slipping) an inclined plane making an angle

\theta

with the horizontal. Then its acceleration is

A

{{g\,\sin \theta } \over {1 - M{R^2}/I}}

B

{{g\,\sin \theta } \over {1 + I/M{R^2}}}

C

{{g\,\sin \theta } \over {1 + M{R^2}/I}}

D

{{g\,\sin \theta } \over {1 - I/M{R^2}}}

Correct Answer

Option B

Solution

A uniform body of radius R, mass M and moment of inertia

I

rolls down (without slipping) an inclined plane making an angle θ with the horizontal. Then its acceleration is

a = {{g\,\sin \,\theta } \over {1 + {I \over {M{R^2}}}}}